http://acm.split.hdu.edu.cn/showproblem.php?pid=6165
題意:給出一個有向圖,要求圖上任意兩點是可達的(不需要互相可達)
題解:dfs,每次搜尋每個結點的兒子然後标記連通即可。
代碼:
#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define d(a) cout<<a<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
const int N = 1000 + 5;
const int mod = 1000000000 + 7;
const double eps = 1e-8;
int dp[N][N];//可達标記
bool vis[N];
vector<int> son[N];
void dfs(int u,int fa){
dp[fa][u]=vis[u]=1;//記為可達
for(int i=0;i<son[u].size();i++){
int v=son[u][i];
if(!vis[v]){
dfs(v,fa);
}
}
}
int main(){
int t,m,n,u,v;
bool flag;
scanf("%d",&t);
while(t--){
flag=1,mem(dp,0),mem(vis,0);
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++){
son[i].clear();
}
for(int i=1;i<=m;i++){
scanf("%d%d",&u,&v);
dp[u][v]=1;
son[u].push_back(v);
}
for(int i=1;i<=n;i++){
vis[i]=1;
dfs(i,i);
mem(vis,0);
}
for(int i=1;i<=n;i++){
for(int j=1;j<i;j++){
if(dp[i][j]||dp[j][i]) continue;
flag=0;
}
}
if(flag){
puts("I love you my love and our love save us!");
}
else{
puts("Light my fire!");
}
}
return 0;
}